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Câu hỏi: Họ nguyên hàm của hàm số $f\left( x \right)=3x\left( x+\cos 3x \right)$ là:
A. ${{x}^{3}}+x\sin 3x-\dfrac{\cos 3x}{3}+C$.
B. ${{x}^{3}}-x\sin 3x-\dfrac{\cos 3x}{3}+C$.
C. ${{x}^{3}}+x\sin 3x+\dfrac{\cos 3x}{3}+C$.
D. ${{x}^{3}}+x\sin 3x+\cos 3x+C$.
Ta có: $I=\int{f\left( x \right)}dx=\int{3x\left( x+\cos 3x \right)}dx={{x}^{3}}+\int{3x.\cos 3x.dx}+C.$
Tính $I'=\int{3x.\cos 3x.dx}$. Đặt $\left\{ \begin{aligned}
& u=3x \\
& dv=\cos 3xdx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=3.dx \\
& dv=\dfrac{1}{3}.\sin 3xdx \\
\end{aligned} \right.$.
$I'=x\sin 3x-\int{\sin 3xdx}=x\sin 3x+\dfrac{1}{3}\cos 3x+C'.$
Vậy $I={{x}^{3}}+x\sin 3x+\dfrac{1}{3}\cos 3x+C.$
A. ${{x}^{3}}+x\sin 3x-\dfrac{\cos 3x}{3}+C$.
B. ${{x}^{3}}-x\sin 3x-\dfrac{\cos 3x}{3}+C$.
C. ${{x}^{3}}+x\sin 3x+\dfrac{\cos 3x}{3}+C$.
D. ${{x}^{3}}+x\sin 3x+\cos 3x+C$.
Ta có: $I=\int{f\left( x \right)}dx=\int{3x\left( x+\cos 3x \right)}dx={{x}^{3}}+\int{3x.\cos 3x.dx}+C.$
Tính $I'=\int{3x.\cos 3x.dx}$. Đặt $\left\{ \begin{aligned}
& u=3x \\
& dv=\cos 3xdx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=3.dx \\
& dv=\dfrac{1}{3}.\sin 3xdx \\
\end{aligned} \right.$.
$I'=x\sin 3x-\int{\sin 3xdx}=x\sin 3x+\dfrac{1}{3}\cos 3x+C'.$
Vậy $I={{x}^{3}}+x\sin 3x+\dfrac{1}{3}\cos 3x+C.$
Đáp án C.